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Transcript
Fibonacci Numbers in Daily Life
Yining Lin, Wangchun Peng, Huixia Chen, Ya Liu
Abstract
Fibonacci sequence was first introduced by an Italian mathematician Fibonacci in his
book Liber Abaci. People named this sequence with Fibonacci’s name because it is
closely connected with nature, science and real life. Therefore Fibonacci sequence is
widely used in many fields and well-worth people exploring. In this paper, we will
introduce two properties of Fibonacci hidden in a game, and further application of
Fibonacci number mainly in botany and composition of pictures.
Key Words: Fibonacci sequence; golden ratio; application.
1 Introduction
Fibonacci sequence was first introduced by an outstanding Italian mathematician
Fibonacci (Leonardo Pisano.1175-1250) in his book Liber Abaci (1202). For he was
born in the commercial center of Pisano, he was also called Leonardo of Pisano [1].
When Fibonacci was a little child, he traveled with his father who worked as a
businessman. He also traveled to other countries like Egypt, Greek and Syria to study
mathematic knowledge from the east when he grew up.
Fibonacci sequence begins with π‘Ž0 = 0, π‘Ž1 = 1, and each subsequent number is the
sum of the previous two numbers, so the recursion formula of Fibonacci sequence
is π‘Žπ‘›+1 = π‘Žπ‘› + π‘Žπ‘›βˆ’1 . People thought the sequence contains the secret of nature, so
they named it with the mathematician’s name Fibonacci. In fact, Fibonacci sequence
closely connects with nature, science and real life. Therefore it is widely used in many
fields and well-worth people exploring.
2 Game about Fibonacci number
2.1 Game
Cut a 8×8 square into four parts( as figure 1 shows), and rearrange the four parts into
a new 5×13 rectangle as figure 2 shows.
Figure 1 Original square
Figure 2 Rearrange into a new rectangle [2]
1
When we calculate the area of figure 1, we can easily get that the area of the square
equals to 64.However, the area of the new rectangle equals to 65.During this process,
we have not abandoned or added any piece of paper into the new rectangle, so the
area of the original square and the rectangle should equal to each other. In order to
figure out the reason for the area change and the hidden mathematical principle
behind, we picked this game as our topic.
2.2 Mathematic principle behind
We observed the square carefully and found out that the length of the sides of the
square and the rectangle are all Fibonacci numbers. We have
𝑛=6
π‘Žπ‘› = π‘Ž6 = 8, π‘Žπ‘›βˆ’1 = π‘Ž5 = 5 π‘Žπ‘›π‘‘ π‘Žπ‘›+1 = π‘Ž7 = 13 .
In fact, Fibonacci sequence has the following property:
(1)
π‘Žπ‘›+1 π‘Žπ‘›βˆ’1 = π‘Žπ‘› 2 + (βˆ’1)𝑛
In our game, the product of an+1 anβˆ’1 actually represents the area of the rectangle
Srectangler , and an 2 represents the area of the square Ssquare .
Srectangle = a7 a5 = 5 × 13 = 65,
π‘†π‘ π‘žπ‘’π‘Žπ‘Ÿπ‘’ + (βˆ’1)6 = π‘Ž6 2 + (βˆ’1)6 == 64 + (βˆ’1)6 = 65.
If we observe the new rectangle carefully, we will find there is a gap on the rectangle.
So the area of the original square has never changed. The area change only caused by
adding extra part into the area of the rectangle.
We use the mathematical induction method to prove this property:
Proof: Step 1 Let n = 1, π‘Ž0 = 0, π‘Ž1 = 1, π‘Ž1 = 1.
Then equality (1)holds.
Step 2 Suppose it is true for n=k. Then we get the equality
π‘Žπ‘˜βˆ’1 π‘Žπ‘˜+1 = π‘Žπ‘˜ 2 + (βˆ’1)π‘˜
Step 3 Now we show it is true for n=k+1.
According to the definition if the Fibonacci sequence, we have the
following recursion formula:
π‘Žπ‘˜+1 = π‘Žπ‘˜ + π‘Žπ‘˜βˆ’1 ;
π‘Žπ‘˜+2 = π‘Žπ‘˜+1 + π‘Žπ‘˜ = 2π‘Žπ‘˜ + π‘Žπ‘˜βˆ’1
Then we get
π‘Žπ‘˜βˆ’1 (π‘Žπ‘˜ + π‘Žπ‘˜βˆ’1 ) = π‘Žπ‘˜ 2 + (βˆ’1)π‘˜
π‘Žπ‘˜ 2 βˆ’ π‘Žπ‘˜βˆ’1 π‘Žπ‘˜ βˆ’ π‘Žπ‘˜βˆ’1 2 = (βˆ’1)π‘˜+1
π‘Žπ‘˜ (2π‘Žπ‘˜ + π‘Žπ‘˜βˆ’1 ) = (π‘Žπ‘˜ + π‘Žπ‘˜βˆ’1 )2 + (βˆ’1)π‘˜+1
Thus
2
(2)
π‘Žπ‘˜ π‘Žπ‘˜+2 = π‘Žπ‘˜+1 2 + (βˆ’1)π‘˜+1
It is true for n=k+1.
Thus property(1) is proved.
(2)
2.3 Another situation
When we reshape the four parts of the original square in another way as figure 3
shows, we get a new polygon. The intersection in the new polygon changes its area
into 63. We can also use another property of Fibonacci sequence to explain why the
change happens.
Figure 3 Rearrange four pars into a new polygon
The Fibonacci sequence also has the following property [3]:
4π‘Žπ‘›βˆ’1 π‘Žπ‘›βˆ’2 + π‘Žπ‘›βˆ’2 π‘Žπ‘›βˆ’4 = π‘Žπ‘› 2 + (βˆ’1)π‘›βˆ’1
Proof For we have (2) π‘Žπ‘› π‘Žπ‘›+2 = π‘Žπ‘›+1 2 + (βˆ’1)𝑛+1
4π‘Žπ‘›βˆ’1 π‘Žπ‘›βˆ’2 + π‘Žπ‘›βˆ’2 π‘Žπ‘›βˆ’4 = 4π‘Žπ‘›βˆ’1 π‘Žπ‘›βˆ’2 + π‘Žπ‘›βˆ’2 (2π‘Žπ‘›βˆ’2 βˆ’ π‘Žπ‘›βˆ’1 )
= π‘Žπ‘›βˆ’2 (3π‘Žπ‘›βˆ’1 + 2π‘Žπ‘›βˆ’2 )
= π‘Žπ‘›βˆ’2 (π‘Žπ‘› + π‘Žπ‘›βˆ’2 + 2π‘Žπ‘›βˆ’1 )
= π‘Žπ‘›βˆ’1 2 + (βˆ’1)π‘›βˆ’1 + π‘Žπ‘›βˆ’2 2 + 2π‘Žπ‘›βˆ’2 π‘Žπ‘›βˆ’1
= π‘Žπ‘› 2 + (βˆ’1)π‘›βˆ’1
Thus the property is proved.
When we calculate the area of the new polygon, we actually do not involve the
intersection part. So the area of the new polygon we get is smaller than the original
square.
2.4 Perfect situation
After we had researched two situations above, it is reasonable to think about the
question that how can we cut the original square to get a perfect rectangle.
Suppose the length of the sides are x and y as figure 4 shows, and we can get a new
rectangle without any gap or intersection.
3
y
x
x+y
x
y
x
y
x
x
Figure 4 The situation we can get a perfect rectangle
Suppose the area of the original square is π‘†π‘†π‘žπ‘’π‘Žπ‘Ÿπ‘’ and the area of rectangle is
π‘†π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ .Then we have
π‘†π‘†π‘žπ‘’π‘Žπ‘Ÿπ‘’ = (π‘₯ + 𝑦)2
π‘†π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’ = (2π‘₯ + 𝑦)π‘₯
Let
π‘†π‘†π‘žπ‘’π‘Žπ‘Ÿπ‘’ = π‘†π‘Ÿπ‘’π‘π‘‘π‘Žπ‘›π‘”π‘™π‘’
Then we have
(π‘₯ + 𝑦)2 = (2π‘₯ + 𝑦)π‘₯
i.e.
π‘₯ 2
π‘₯
( ) βˆ’( )βˆ’1= 0
𝑦
𝑦
Thus we get the solution
π‘₯ 1 ± √5
=
𝑦
2
Because x and y are the length of the sides, so we only take the positive one. It is easy
to realize that
𝑦
2
=
β‰ˆ 0.618
π‘₯ 1 + √5
In fact, if y equals to π‘Žπ‘›βˆ’2 and x equals to π‘Žπ‘›βˆ’1 , the proportion of x for y represents
the proportion of π‘Žπ‘›βˆ’1 for π‘Žπ‘›βˆ’2 . When n tends to infinity, the ratio tends to be 0.618
(golden ratio).So the perfect situation above happened when x equals to π‘Žπ‘›βˆ’1 , y
equals to π‘Žπ‘›βˆ’2 and n tends to be infinity.
2.5 Fibonacci numbers and golden ratio
From the result above, we find that Fibonacci numbers are close connected with
golden ratio. Therefore in the following part, we will apply the continued-fraction
approximation to show the connection between them.
The golden ratio is very useful in Aesthetics and the Optimization, for example, the
golden rectangle and the golden ratio method. But in some cases, the golden ratio
method can’t be used directly, for example, when we deal with the Optimization
problem whose domain is a set of discrete points, which means the total length of the
4
domain can be computed as an integer, but the domain is discrete.
So find some fractions close to the golden ratio would be the solution for the problem
mentioned above.
The following listed is some conclusion we get by analyzing the problem.
Property 1:
Obviously, if we decrease the error of the approximation, the
denominator of the fraction will be increased. But for a concrete discrete
Optimization problem, its domain length is a concrete integer, and only
when the denominator of the fraction equals that integer can that fraction
be used to solve the problem.
Corollary 1:
So we need fractions with different denominators to solve the discrete
Optimization problem with different domain length.
Property 2:
However, when the denominator of the fraction less than the domain
length of the problem, we can also find a way to use that fraction to
solve the problem. The idea is simply by dividing the domain of the
problem into several parts and the length of each part is the denominator
of the fraction except the last part whose length is less than it. Then we
can use the fraction to search the optimum part by part, but for the last
part we need use dichotomization to find the optimum of this part.
Finally, by simply comparing all the optimums and choosing the best
one of them, we can get the problem solved.
Corollary 2:
The best fraction approximating the golden ratio varies with the change
of the domain length of the problem.
Definition 1:
We define the best fraction approximating the golden ratio as G[n],
when we limit the denominator of the fraction less than an integer n. So
the denominator of G[n] is less than n, and among all fractions with
denominator less than n, G[n] is the best one to approximate the golden
ratio.
Till now we have get all the preliminaries prepared to show the connection between
the Fibonacci sequence and the golden ratio.
What we want to show below is an equation:
F[k βˆ’ 1]
G[n] =
π‘€β„Žπ‘’π‘› 𝐹[π‘˜ βˆ’ 1] < 𝑛 ≀ F[k]
F[k]
Where the sequence denoted by {F[k]} is the Fibonacci sequence.
Limited by space, we put the detailed proof of the above equation in the appendix.
5
3 Further applications
3.1 Application in botany
Figure 5 The Fibonacci spiral appeared in some kind of aloes
Figure 6 Flowers whose numbers of petals are Fibonacci numbers
Considering the number of the petals of the flowers, some of them follow Fibonacci
numbers. One possible reason for why this phenomenon happens is that they try to
decrease the overlapped area to get more sunlight.
Figure 7 The hidden Fibonacci spiral in sunflower
6
According to figure 7, when we count the number of its spirals, we will find there are
13 counterclockwise spirals and 21 clockwise spirals. These two numbers are both
Fibonacci numbers. One possible reason for the Fibonacci spiral appears in the
sunflower is to fully use the space.
Figure 8 The number of branches follows the Fibonacci numbers
Botanists also find the numbers of branches of trees are always Fibonacci number.
They find that after a certain period of time, each old branch of a tree will get a new
one and need one more period of time to turn to an old one. If the tree has only one
branch at very beginning, after a year it will have two branches and in next circle, it
has three branches. Every year the total branches of the tree composed a Fibonacci
sequence. One possible reason for it is that every new branch needs one year to get
mature.
Since Fibonacci numbers shows in some of the flower petal numbers and the
arrangements of the inflorescence, many plants do not show any Fibonacci number in
their arrangements [4]. In fact, scientists have not figured out the pattern accurately the
plants arrange their leaves and flowers. We can only assume that plants apply
Fibonacci numbers and the golden ratio in their growth to make their space
utilizations more effectively. However, it is also possible that plants do not understand
Fibonacci sequence at all, and maybe they just grow to the shape they like.
3.2 Application in composition
3.2.1Fibonacci spiral
One of the most common applications of Fibonacci sequence in the composition is
Fibonacci spiral. As figure 5 shows, the sides of the square are all different Fibonacci
numbers.
Figure 5 Fibonacci spiral
7
The following pictures show the hidden Fibonacci spirals in composition.
Figure 6 Picture of rose
The photographer use the spiral into composition to highlight the rose in figure
6.Figure 6 shows the hidden Fibonacci spiral in the picture.
Figure 7 Stage photo of Sherlock
Figure 7 is a stage photo picked from BBC’s show Sherlock. The starting point of
Fibonacci spiral in Figure 7 highlights the girl’s eye, and the arm of the man help
form the spiral. The application of Fibonacci spiral can make the picture looked more
harmonious [5].
3.2.2 The Golden Triangle
The triangle ABC and triangle CBX are both isosceles triangles and similar to each
other. The waist and the bottom edge length ratio for 1.618. Figure 8 shows how a
golden triangle looks like.
Figure 8 The Golden Triangle
8
Figure 9 Golden triangle used in stage photos
Figure 9 shows the hidden golden triangle being applied in two stage photos. The
application of the golden triangle also brings the sense of beauty to the pictures.
3.3 Application in bees
The breed form of bees is very special, for the male bees are all hatched from
unfertilized eggs and the female ones are all hatched from fertilized one. Hence the
male bees are only the fathers of the female ones, and male bees also have no father at
all. When scientists observed the family tree of the bees, they found that the number
𝐺𝑛 of the n generation of the first male bee equals to the π‘Žπ‘› in Fibonacci sequence.
Figure 10 shows the family tree of a male bee, the symbol♀represents female bees
and the symbolβ™‚represents the male bees.
Figure 10 The family tree of a male bee [6]
Reference
[1] Hongwei Gao. Fibonacci in Eveywhere. Journal of Yulin college. 2002,12(4)
[2] Feifei Jia. The research and the application of Fibonacci sequence. Technology
Innovation a. 2014(13).
[3] Zhenkui Wu. Fibonacci Sequence Appreciation. Harbin institute of technology
press.2012.
[4]. Jun Shi. Is there any hidden the secret numbers in botany. Studies for middle
school students.2012(7).
[5] Fomalhaut. Learn how to use golden ratio to take the photo through watching the
show Sherlock. http://academy.fengniao.com/429/4299797.html. 2015.06.21.
[6] About Fibonacci sequence. http://m.doc88.com.2015.06.22
9
Appendix
The proof for the equation:
F[k βˆ’ 1]
π‘€β„Žπ‘’π‘› 𝐹[π‘˜ βˆ’ 1] < 𝑛 ≀ F[k]
F[k]
Where the sequence denoted by {F[k]} is the Fibonacci sequence.
G[n] =
If we denote the golden ratio by a label x, then according to the definition of the
golden ratio we get the label x satisfying the following equation.
1βˆ’x
=x
x
Where the variable denoted by x belong to the interval [0, 1].
Now we use elementary mathematics to compute G[n].
By solving the equation above we get
x β‰ˆ 0.61803398874989484820458683436564
1
1
=
1
1
1
1
floor
(
)
+
[
βˆ’
floor
(
x
x
x
x )]
1
1
0 ≀ βˆ’ floor ( ) < 1
x
x
The label floor(x) represents the greatest integer less or equal to the number x.
x=
1
Obviously, floor (x) > 1, we can neglect
1
x
1
βˆ’ floor (x) and get a approximation of
x, which mean:
xβ‰ˆ
1
1
floor ( x )
Next we will prove:
1
1
G [floor ( )] =
1
x
floor (x )
Proof:
1
q
1
1
If G [floor (x)] = p, then 1 ≀ q ≀ p ≀ floor (x) ≀ x and we need
reach its minimum.
q
p
βˆ’x
q
qβˆ’pβ€’x
βˆ’x=
p
p
Based on the given conditions, we get p β€’ x ≀ 1, so by letting the numerator reaching
its minimum and letting the denominator reaching its maximum we get:
1
q=1 ,
p = floor ( )
x
then we proved that:
10
1
1
G [floor ( )] =
1
x
floor (x )
Obviously,
G[1] = 1
To find the next fraction with a bigger denominator and higher accuracy, we can
simply use the method above to find a fraction to approximate the numberβ€”β€”
1
x
1
βˆ’ floor (x) .
1
1
1
1βˆ’x
[ βˆ’ floor ( )] = βˆ’ 1 =
=x
x
x
x
x
So the method to compute G[1] can be iteratively used to compute G[n] and then we
can get a sequence which is a subset of {G[n]} (we neglect the detailed proof for this
assertion here, but it can be proved by using the mathematical induction) . What’s
more, if we denote the sequence by {x[n]}, then we can get a recursive relation:
1
x[n] = 1+x[nβˆ’1] , x[0] = G[1] = 1.
F[n]
It would be easy to show the formulaβ€”β€”x[n] = F[n+1] .
We neglect the proof of the set {G[n]} is equal to the set {x[n]}, because the proof
need to use many properties about the continued-fraction approximation theory. If you
feel interested in it you can consult Wikipedia’s entry on continued-fraction. Actually
the continued-fraction approximation method can also be applied to solve the problem
exists in the twelve-tone equal temperament.
At last, by simply applying the theory about the Fixed-Point Iteration we can verify
that {x[n]} converges to the golden ratio x.
11
Note
In fact, Fibonacci number is very interesting and we can see it in many objects.
Especially in nature, such as the sorting of the petals, and leaves, they are consistent
with Fibonacci numbers. Fibonacci also exist in insects such as the honeycomb.
--By Huixia Chen
Through this class and by doing the group assignment of it, I learned that life is the
source and impetus of the development of mathematics. So keeping a vigorous eye on
the problems appeared in real life and solving them by using mathematical idea will
certainly give you some ideas to develop new mathematics.
--By Wangchun Peng
In this course, we actually learn more about β€˜daily life’ than β€˜mathematic’. We did not
do the endless calculation, did not prove endless theorems and did not solve endless
equations. What we did was finding, learning and improving. During this whole week,
we work hard to find the mathematics hidden in our daily lives, learn from other
students from our class, and never stop trying to improve ourselves.
From the team work I learnt how to communicate with others, how to work efficiently
and how to be patient. From the presentation I learnt how to consider problems in
different ways, how to give a better presentation, and to respect the labors of others.
I used to feel cool when I understood the math theory that others could not understand
at all. But I gradually realized that people do not care what you can do only in
mathematic field, but they care about what you can do to make the world better. There
are several name of mathematicians in my text book my mum has never heard about.
But people always remember the ones who devoted themselves to the society even
though the ones who did not get high education. I do not expect to let people
remember me, but I want to make something in the world better.
When I sat down and looked back the whole week, I got the thinking above. I hope I
can keep the idea in my mind: never stop trying figure out things, never stop
improving ourselves and never forget to take others’ shoes.
--By Yining Lin
I think the idea of the course Mathematics in Daily Life is to provide more learning
methods and thinking to my math study. This course also helps me solving the
problems in our daily lives and stimulates my enthusiasm for English
--By Ya Liu
Work division
Yining Lin : presentation; slide show ; paper writing
Wangchun Peng: presentation; slides show; paper writing
Huixia Chen: presentation; paper writing; data collection
Ya Liu: topic choose; slides show (absent from presentation and low join rate in paper
writing)
12